Ask question

Ask question

All categories

1 year ago

7

Justine is ice-skating at the Lloyd Center what is her final velocity if she accelerates at a rate of 2.0 meters per second for

3.5 seconds

1 answer:

lilavasa [31]1 year ago

7 0

2*3.5 = 7m/s

You multiply the acceleration per the time (they both are in seconds, otherwise, you should set them in the same units).

You might be interested in

A 2 kg stone moves with a speed of 1 m/s. A second 2 kg stone is moving twice as fast. Compare their kinetic energies.

alekssr [168]

D
is the answer
Well it should be

5 0

2 years ago

Read 2 more answers

Sally finds herself stranded on a frozen pond so slippery that she can't stand up or walk on it. To save herself, she throws one

8_murik_8 [283]

Answer:

a) 2.5 m/s. (In the opposite direction to the direction in which she threw the boot).

b) The centre of mass is still at the starting point for both bodies.

c) It'll take Sally 12 s to reach the shore which is 30 m from her starting point.

Explanation:

Linear momentum is conserved.

(mass of boot) × (velocity of boot) + (mass of sally) × (velocity of Sally) = 0

5×30 + 60 × v = 0

v = (-150/60) = -2.5 m/s. (Minus inicates that motion is in the opposite direction to the direction in which she threw the boot).

b) At time t = 10 s,

Sally has travelled 25 m and the boot has travelled 300 m.

Taking the starting point for both bodies as the origin, and Sally's direction as the positive direction.

Centre of mass = [(60)(25) + (5)(-300)]/(60+5)

= 0 m.

The centre of mass is still at the starting point for both bodies.

c) The shore is 30 m away.

Speed = (Distance)/(time)

Time = (Distance)/(speed) = (30/2.5)

Time = 12 s

Hope this Helps!!!

7 0

2 years ago

Read 2 more answers

A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the

Fofino [41]

Answer:

2 x 10⁻³ volts

Explanation:

B = magnetic of magnetic field parallel to the axis of loop = 1 T

$\frac{dA}{dt}$ = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B $\frac{dA}{dt}$

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

7 0

2 years ago

A 92-kg skier is sliding down a ski slope that makes an angle of 30 degrees above the horizontal direction. The coefficient of k

9966 [12]

Answer:

a = 4.05 m/s²

Explanation:

Known data

m= 92 kg : mass of the skier

θ =30° :angle θ of the ski slope with respect to the horizontal direction

μk= 0.10 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the skier

W: Weight of the skier : In vertical direction

N : Normal force : perpendicular to the ski slope

f : Friction force: parallel to the ski slope

Calculated of the W

W= m*g

W= 92kg* 9.8 m/s² = 901,6 N

x-y weight components

Wx= Wsin θ= 901,6 N *sin 30° = 450.8 N

Wy= Wcos θ = 901,6 N *cos 30° =780.8 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay ay = 0

N - Wy = 0

N = Wy

N = 780.8 N

Calculated of the f

f = μk* N= 0.10*780.8 N

f = 78.08 N

We apply the formula (1) to calculated acceleration of the skier:

∑Fx = m*ax , ax= a : acceleration of the block

Wx - f = m*a

450.8- 78.08 = ( 92)*a

372.72 = (92)*a

a = (372.72)/ (92)

a = 4.05 m/s²

6 0

2 years ago

A 60.0-kg mass person wishes to push a 120-kg mass box across a level floor. The coefficient of static friction between the pers

Aneli [31]

Answer:

μ = 0.350

Explanation:

For the person to able to move the box, the force exerted by the person on the box must equal the force exerted by the box:

$F_{p} = F_{b}$

In this case, force can be calculated as a product of mass (m) by the acceleration of gravity (g) and the coefficient of static friction (μ):

$m_{p}*g*\mu_{p}=m_{b}*g*\mu_{b}\\m_{p}*\mu_{p}=m_{b}*\mu_{b}\\60*0.7=120*\mu_{b}\\\mu_{b}= 0.35$

Therefore, for the person to be able to push the box horizontally, the coefficient of static friction between the box and the floor should not be higher than 0.350.

8 0

2 years ago